In general, it is better to approach such a question numerically, since your sum is absolutely convergent. However, in your particular case, it is possible to compute this explicitly without any numerical calculations. Notice that non-zero summands that appear in your sum correspond to squarefree $d$ (otherwise $\mu(d)=0$). Next, take a large $X$ and consider all squarefree $d$ with $2<d\leq X$. If such a $d$ is even, then $d=2d_1$ and $d_1>1$ is squarefree and odd. Since $\varphi(d)=\varphi(d_1)$ is even, we have $[d,\varphi(d)]=[d_1,\varphi(d_1)]$ (here $[a,b]=\mathrm{lcm}(a,b)$). On the other hand, $\mu(d)=\mu(2d_1)=-\mu(d_1)$, therefore
$$
\frac{\mu(d)}{[d,\varphi(d)]}+\frac{\mu(d_1)}{[d_1,\varphi(d_1)]}=0.
$$
Every even squarefree $d$ in $(2,X]$ is a member of one such pair, and same is true for odd squarefree $d$ with $1<d\leq X/2$, so
$$
\sum_{d\leq X}\frac{\mu(d)}{[d,\varphi(d)])}=1-\frac12+\sum_{\text{odd }d \text{ in }(X/2,X]}\frac{\mu(d)}{[d,\varphi(d)]}.
$$
This last summand can be estimated as follows
$$
\left|\sum_{\text{odd }d \text{ in }(X/2,X]}\frac{\mu(d)}{[d,\varphi(d)]}\right|\leq \sum_{\text{odd }d>X/2}\frac{1}{[d,\mathrm{ord}_2(d)]}\ll \exp(-1/3(\ln X\ln\ln X)^{1/2})=o(1),
$$
by the paper you linked. Therefore, your sum is equal to $1/2$.
Let $c:=\gcd(\phi(d),d)$. then setting $r:=dk$ the sum can be rewritten as
$$c\sum_{k\leq x/d} \gcd(\tfrac{\phi(d)}c,k) \approx \frac{c^2x}{d\phi(d)}f(\tfrac{\phi(d)}c),$$
where $f(m) := \sum_{k=1}^m \gcd(k,m)$ is the multiplicative function given by its values on prime powers $f(p^s) = p^{s-1}((p-1)s+p)$. See also OEIS A018804.
Best Answer
Here is a question that addresses the mentioned asymptotics - https://math.stackexchange.com/questions/2683190/showing-sum-n-leq-x-frac1-phi-n-c-log-x-o1?noredirect=1&lq=1 Roughly, you can proceed by the standard convolution method via approximating $1/\varphi(n)$ as $1/n$.